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26% (02:41) correct
74%
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wrong
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Is 1/(x - y) < y - x
(1) 1/x < 1/y
(2) 2x = 3y
(1) 1/x < 1/y
(2) 2x = 3y
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
07 Jan 2011, 09:37
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Proposition 1:
1/X < 1/Y
This is equal to X > Y | x, y both > 0 or both < 0
Case 1: Both > 0. X > Y, therefore X-Y >0 , Y-X <0, so 1/(X-Y) > 0 and Y-X<0 so FALSE
Case 2: X < 0, Y>0. Cross multiple and switch the signs, you get Y>X. There for Y-X>0, X-Y<0.
1/(X-Y)<0, Y-X>0, so case 2, 1/(X-Y) < (Y-X) is TRUE
Therefore statement 1 is insufficient by contradiction.
Proposition 2:
2X=3Y
Case 1: X=3, Y=2 - therefore X-Y=1, Y-X = -1
1/(X-Y) < Y-X ----> 1/1 < -1 = FALSE
Case 2: X=-3, Y=-2 - therefore X-Y=-1, Y-X = 1
1/(X-Y) < Y-X ----> 1/-1 < 1 = TRUE
Insufficient by contradicition
Both Statments together - Intutitively, neither statement removes the ambiguity of the sign of X or Y. I can't think of a proof.
Therefore answer is E.
1/X < 1/Y
This is equal to X > Y | x, y both > 0 or both < 0
Case 1: Both > 0. X > Y, therefore X-Y >0 , Y-X <0, so 1/(X-Y) > 0 and Y-X<0 so FALSE
Case 2: X < 0, Y>0. Cross multiple and switch the signs, you get Y>X. There for Y-X>0, X-Y<0.
1/(X-Y)<0, Y-X>0, so case 2, 1/(X-Y) < (Y-X) is TRUE
Therefore statement 1 is insufficient by contradiction.
Proposition 2:
2X=3Y
Case 1: X=3, Y=2 - therefore X-Y=1, Y-X = -1
1/(X-Y) < Y-X ----> 1/1 < -1 = FALSE
Case 2: X=-3, Y=-2 - therefore X-Y=-1, Y-X = 1
1/(X-Y) < Y-X ----> 1/-1 < 1 = TRUE
Insufficient by contradicition
Both Statments together - Intutitively, neither statement removes the ambiguity of the sign of X or Y. I can't think of a proof.
Therefore answer is E.
07 Jan 2011, 10:15
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Is \(\frac{1}{x - y }< y - x\)
Since the numerator (\(1+(x-y)^2\)) is always positive then the question basically becomes whether denominator (\(x-y\)) is negative:
(1) \(\frac{1}{x}<\frac{1}{y}\).
If both unknowns are positive or both unknowns are negative then \(y<x\) (if both are positive, cross multiply to get \(y<x\) and if both are negative, cross multiply and flip the sigh twice to get \(y<x\) again) and the answer will be NO but if \(x<0<y\) given inequality also holds true and in this case the answer will be YES (if \(x\) is any negative number and \(y\) is any positive number then \((\frac{1}{x}=negative)<(positive=\frac{1}{y})\)). Not sufficient.
(2) \(2x=3y\).
The above means that \(x\) and \(y\) have the same sign, next: \(\frac{x}{y}=\frac{3}{2}\).
If both \(x\) and \(y\) are positive (for example 3 and 2 respectively) then \(0<y<x\) and the answer will be NO but if both \(x\) and \(y\) are negative (for example -3 and -2 respectively) then \(x<y<0\) and the answer will be YES. Not sufficient.
(1)+(2) As from (2) \(x\) and \(y\) have the same sign then from (1) \(y<x\) and the answer to the question is NO. Sufficient.
Answer: C.
- Is \(\frac{1}{x-y}<y-x\)?
Is \(\frac{1}{x-y}+x-y<0\)?
Is \(\frac{1+(x-y)^2}{x-y}<0\)?
Since the numerator (\(1+(x-y)^2\)) is always positive then the question basically becomes whether denominator (\(x-y\)) is negative:
- Is \(x-y<0\)? (or is \(x<y\)?)
(1) \(\frac{1}{x}<\frac{1}{y}\).
If both unknowns are positive or both unknowns are negative then \(y<x\) (if both are positive, cross multiply to get \(y<x\) and if both are negative, cross multiply and flip the sigh twice to get \(y<x\) again) and the answer will be NO but if \(x<0<y\) given inequality also holds true and in this case the answer will be YES (if \(x\) is any negative number and \(y\) is any positive number then \((\frac{1}{x}=negative)<(positive=\frac{1}{y})\)). Not sufficient.
(2) \(2x=3y\).
The above means that \(x\) and \(y\) have the same sign, next: \(\frac{x}{y}=\frac{3}{2}\).
If both \(x\) and \(y\) are positive (for example 3 and 2 respectively) then \(0<y<x\) and the answer will be NO but if both \(x\) and \(y\) are negative (for example -3 and -2 respectively) then \(x<y<0\) and the answer will be YES. Not sufficient.
(1)+(2) As from (2) \(x\) and \(y\) have the same sign then from (1) \(y<x\) and the answer to the question is NO. Sufficient.
Answer: C.
